# What is the volume of #2.0 # #mol# #dm^-3 # sodium chloride needed to prepare #250cm^3# of #0.2 ## mol# #dm^-3# sodium chloride solution ?

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What are the steps to solve this ?

What are the steps to solve this ?

##### 1 Answer

#### Answer:

#### Explanation:

Start by calculating the ratio that exists between the concentration of the **stock solution** and the concentration of the **diluted solution**.

#"DF" = (2.0 color(red)(cancel(color(black)("mol dm"^(-3)))))/(0.2color(red)(cancel(color(black)("mol dm"^(-3))))) = color(blue)(10)#

This ratio gives you the **dilution factor**,

Now, the thing to remember about the dilution factor is that can also be calculated by taking the ratio that exists between the volume of the **diluted solution** and the volume of the **stock solution**.

#"DF" = V_"diluted"/V_"stock"#

In your case, the diluted solution has a volume of

#V_"stock" = "250 cm"^3/color(blue)(10) = color(darkgreen)(ul(color(black)("25 cm"^3)))#

I'll leave the answer rounded to two **sig figs**, but keep in mind that you have one significant figure for the concentration of the diluted solution.

So, in order to prepare this solution, you need to take *enough water* to get the total volume of the solution to